Chapter 2 Exercise 2A

Transcription

1 Chapter Eercise A Q. 1. (i) u 0, v 10, t 5, a? a a m/s (ii) u 0, a, t 5, s? s ut + at s (0)(5) + ()(5) 5 m Q.. (i) u 0, v 4, a 3, t? t t 8 s (ii) u 0, a 3, t 8, s? s ut + at s (0)(8) + (3)(64) 96 m Q. 3. u 0, a 3, s 6, v? v u + as v 0 + (3)(6) v 6 m/s Q. 4. u 50, v 70, s 300, a? v u + as 4,900,500 + (a)(300) a 4 m/s u 50, v 70, a 4, t? t t 5 s Q. 5. a 0.5, s 600, t 40, u? s ut + at 600 u(40) + (0.5)(1,600) u u u 5 m/s Q. 6. u 3, v 11, t 6, s? s ( u + v ) t s ( ) (6) 4 m Q. 7. u 3, v 0, s 6, a? v u + as a(6) a 3 4 m/s u 3, v 0, a 3 4, t? ( 3 4 ) t t 4 s Q. 8. (i) u 70, v 50, t 8, a? a(8) a m/s u 70, t 8, a, s? s ut + at s 70(8) + ( ) (64) s (ii) u 50, v 0, a, s? v u + as 0,500 5s s 500 m Q. 9. (i) u 4, v 0, a 8, s? v u + as ( 8)s s 36 m (ii) v u + as 0,304 + ( 8)s s 144 m Q. 10. (i) 7 km/hr 7,000 m 3,600 s 0 m/s (ii) 48 km/hr 48,000 m 3,600 s v u + as 1, a(500) 9 a 9 m/s ( 9 ) t t 30 s 40 3 m/s 1

2 (iii) v u + as 0 1, ( 9 ) s s 400 m Q. 11. (i) 1 km/hr 1,000 m 3,600 s 5 18 m/s 7 km/hr 0 m/s ad 54 km/hr 15 m/s v u + as (a)(35) a m/s (ii) v u + as ( ) s s 45 m Q. 1. Let t be the time of meetig. s 1 5t + (3)t 5t + 3 t s 7t + ()t 7t + t s 1 + s 16 1t + 5 t 16 t 6 ( t 54 5 rejected ) At t 6, v 1 u + at 5 + (3)(6) 3 m/s v 7 + ()(6) 19 m/s Q. 13. (i) They will meet whe s 1 + s t + (4)t + 7t + ()t t + 3t 400 3t + 10t (3t + 40)(t 10) 0 t 40 OR t 10 3 t 10 s (t 40 is rejected as impossible) 3 (ii) s 1 3(10) + (4)(10) 30 m s 7(10) + ()(10) 170 m Q. 14. (i) 30 km/hr 30,000 m 3,600 s t t 4 6 s (ii) v u + as ( 50 3 ) 0 + ()s s 65 9 m (iii) v u + as Eercise B 0 ( 50 3 ) + (a)() a 65 9 m/s Q. 1. (i) u 0, a, v t t 15 s (ii) Speed (m/s) 30 A B 5 3 m/s Time (s) Total distace covered Area uder graph A + B + C (15)(30)+ (1)(30) + (10)(30) m (iii) u 30, v 0, t a 10a 30 a 3 m/s Magitude of deceleratio is 3 m/s Q.. (i) a(5) a 4 m/s C

3 Q. 3. (i) (ii) s ut + at s 0(5) + (4)(5) 50 m (iii) Remaiig distace 00 m. 00 m at 0 m/s takes 10 secods. The total time take s Speed (m/s) 50 (i) Durig deceleratio Area uder graph 150 Area C 150 (t)(60) t 150 t 5 s (ii) Durig acceleratio u 0, v 60, a 3 Q Durig acceleratio u 0, v 50, a t t 10 s Durig deceleratio u 50, v 0, a t 10t 50 t 5 s Time (s) Total distace travelled Area uder graph (10)(50) + (5)(50) + (5)(50) , ,65 m Total distace (ii) Average speed Total time 1, Speed (m/s) 60 1, m/s A B C t t 0 s Total distace covered Area uder graph (0)(60) + (75)(60) , ,50 m (iii) Average speed 5, Q. 5. (i) v u + as (40) 0 + ()s s 400 m t t 0 s (ii) v u + as 0 (40) + ( 5)s s 160 m ( 5)t t 8 s 5.5 m/s (iii) Remaider 1, m Q. 6. (i) 440 m at 40 m/s takes 11 secods. Total time s a(9) 75 t Time (s) a 3 m/s 3

4 (ii) v u + as 0 (7) + (a)(54) a m/s (iii) d part: ( ) t t 4 s Distace Area uder the curve (13)(7) 175 m Average speed Total distace Total time Q. 8. Q (a)(4) 10 a m/s t 1 t Let t time spet acceleratig (t)(10) + (1 t)(10) 100 5t t 100 a 10.5 m/s t t 4 (iv) First part: t t 5 s d part: m/s Q. 10. t 18 t (t)(5) + (18 t)(5) t t a t t 8 s d 8 4 Q ( ) t t s Aswer: After 5 secods ad after ( ) s B A C 40 Area uder the curve 980 ()(10) + (15)() + (40 )(5) s y z 40 0 z Area 700 Q. 11. (i) 40 (0)(40) + 40y + (5)(40) y Time s a 60 y y 5 z 3a

5 a m/s Sice 60a 40 3az z 60a 3a s Area 8,800 (60)(40) + 40y + (0)(40) 8,800 1, y ,800 40y 7,00 Distaces are 1,00, 7,00, 400 metres (ii) 40 y 180 Total time s a b c d Shaded regio 1 km 1,000 m The deceleratio 3a 3 ( 3 ) m/s (from (i)) d 0 10 s Shaded regio 1,000 0c + (d)(0) 1,000 0c + (10)(0) 1,000 0c ,000 0c 900 c 45 Dotted regio: u 40, v 0, a, t b 0 40 b b 10 Area 10(0) + (10)(0) 300 m Total area 8,800 (60)(40) + 40a ,800 40a 6,300 a Total time Eercise C Q Etra time t t 5 s.5 secods more tha the first time QED 5 t 5 Area uder the curve 100 (5)(10) + (t 5)(10) 100 t 1.5 s Q.. (i) v 0 + ( 1 ) (15) m/s 0 + ( 5)t t 4 s (ii) Distace Area uder the curve Q. 3. (i) t t 1 s s ( u + v ) t 48 ( ) t t 4 s ( 19 ) ( ) m 5

6 (ii) First part: a ut + at 0(1) + ()(1) 144 m Total distace m Total time s Average speed 19 1 m/s 16 Q. 4. s 1 ut + at 0(t) + (4)t t s 0t s ut + at 48 u(4) + (a)(16) u + a 1 (1st ad d parts) Solvig these gives (i) a 3 m/s (ii) u 6 m/s (iii) First 6 secods: s ut + at s (6)(6) + (3)(36) 90 m s 1 s t 0t The distace travelled m t 10 s s 00 m Q. 8. O A 6s B s C 105 m 63 m Q. 5. (i) v t; v 0 + t (ii) s 1 10t + 3 t ; s 0t + t (iii) v 1 v t 0 + t t 10 s (i) A to B: u u, s 105, t 6, a a s ut + at 105 6u + 18a A to C: u u, s 168, t 8, a a s ut + at Q. 6. (iv) s 1 s 10t + 3 t 0t + t t 0 OR t 0 s A 4s B 6s C 48 m 10 m A to B: u u, t 4, s 48, a a 168 8a + 3a Solvig gives a 3.5, u 7 Aswer: a 3.5 m/s (ii) O to A: u 0, v 7, a 3.5, s s v u + as s s 7 m 6 s ut + at 48 4u + 8a u + a 1 A to C: u u, t 10, s 150, a a s ut + at u + 50a u + 5a 15 Solvig gives a 1, u 10 Aswer: 1 m/s Q. 7. (i) s ut + at 18 u() + (a)(4) u + a 9 (1st part) Q. 9. 1st part: s ut + at 39 u(1) + a(1) u + a 78 1st ad d parts: 76 u() + a() u + a 76 First three parts: 111 u(3) + a(3) 6u + 9a u + 3a 74 Solvig the first two equatios gives a, u 40 Solvig the last two equatios gives a, u 40 They are cosistet.

7 Stoppig: v u + as 0 (40) + ( )s s 400 m It will travel a further m Q. 13. Speed (m/s) Q. 10. a to b: s ut + at Q u(5) + (a)(5) u + 5a 8 a to c: 40 u(8) + a(8) u + 8a 10 Solvig these gives u 7 3 m/s, a 3 m/s a to d: a ut + at t + ( 3 ) t t + 7t (use formula) t 10.4 s ( 17.4 is rejected). The time take s Area (10) (6) 1,000 5 m/s Distaces are: (10)(5) 15; 3(5) 800; (6)(5) 75 m Q. 1. (i) : t 3 : : 4 t T Let the top speed v v t v 7 The time take 90 secods T T v + v 7 Q. 14. (i) 7v + v 14 9v 14 9v v 1,60 v 140 m/s Distace (90)(140) 6,300 m 6.3 km Time (s) 3 (0) 15 s 4 t 4 (0) 5 s (ii) (iii) v 0 + (1)(15) 15 m/s s Area uder curve (0)(15) 150 m (ii) Area (1) (8) 1,000 0 m/s T 1 Area 1 Area 14T (1)(0) + 0(T 1) T 0 s s 14T 80 m 7

8 Q st part: s ut + at Q. 16. (i) 4 u() + a() u + a 1 1st ad d parts: 48 u(3) + a(3) u + 3a 3 Solvig these gives a 8, u 4 First three parts: 7 4t + (8)t t + t 18 0 t 3.77 (t 4.77 is rejected) Time take s 16 Q Q. 17. (i) s ( ) t 4 t s 0 + (1)t t Distace apart, s s 1 + s 3 4 t Whe t 10, s 3 (100) 75 m 4 (ii) Whe s t 108 t 144 t 1 s This is 1 10 secods later Q. 18. Let v top speed A B C 16 8 Distace i A (16)(16) 18 m 4 8 (ii) Distace i C (8)(16) 64 m Total distace travelled 19 m Remaider for B m Time take s Total time s 4 ( v 4 + v 8 ) v 1,00 ( 3v 8 ) v 1,00 3v 16 1,00 v 6,400 8 v 80 m/s Time v 4 + v s Q. 19. Let t time after cyclist passes P 8 t T Let T the time take : t d : a : 1 3 : 3 3 T ad t 3 T v 0 + (1) ( 3 T ) 3 T Area 300 (T) ( 3 T ) 300 T 30 s Greatest gap v 1 v (t 5) t 17 at t 17, s 1 1(17) 04 m s (1)(17 5) 7 m Gap m Q. 0. Greatest gap v 1 v 8 + 4t (t ) 8 + 4t t 6 t 16 at t 16, s 1 8(16) + (4)(16) 640 at t 16 14, s 30 (14) + (3)(14) 714 Gap m QED

9 Q. 1. Q.. A G P 1 P After t secods, Alberto has travelled s 1 1t + t After t secods, Gustav has travelled s t At both P 1 ad P, s 1 s + 1t + t t + t 4t t, Aswer: (i) After secods 10 (ii) After 0 secods more 50 6 Area uder the curve 696 (10) (6) a 1 (10) a 1 1. m/s Similarily, a m/s Q. 3. Take speeds, acceleratios, distaces relative to the goods trai. Let p the passeger trai ad g the goods trai. The iitial relative speed, u pg u p u g m/s The relative distace 1,500 m The fial relative speed is zero, sice the two trais must evetually be travellig at the same speed to avoid a crash. v u + as 0 (50) + a(1,500) a 5 6 m/s The relative deceleratio is, therefore, 5 6 m/s The actual deceleratio of the passeger trai is 5 6 m/s, sice the goods trai does ot decelerate at all. Q. 4. (i) Iitial relative speed, u m/s Relative distace 10 m Fial relative speed 0 m/s v u + as 0 (1) + a(10) a 3 5 m/s t : t : 1. 5 : : T, t 3 8 T (ii) (i) u 1 a 1 s t? s ut + at v 0 + (1.) ( 5 8 T ) 3 4 T Area 696 (T ) ( 3 4 T ) 696 T 1,856 T t + ( 1)t t 4t (t 6)(t 18) 0 t 6, 18 Aswer: After 6 secods 9

10 10 (ii) u 1 a 1 s s t t s ut + at s 1t t ds dt 1 t Eercise D 0 (Sice s is a miimum) t 1 At t 1, s 1(1) (1) 7 m This meas that they have travelled a distace of 7 m towards each other, ad so the distace betwee the will be m. Q. 1. (i) s 35t 4.9t 0 t( t) 0 t 0 OR t s (ii) v t 0 s 35 ( 5 t ) 4.9 ( 5 7 ) m Q.. (i) u(1) 4.9(1) 16.1 u u 1 m/s (ii) v t 0 t at t 15 7, s y 1 ( 15 (iii) s y 0 7 ) 4.9 ( 15 7 ) m 1t 4.9t 0 t 0 OR t s Q. 3. (i) s 1 s 0(t) + 4.9t 14.7 (t 1) (t 1) t 3(t 1) + (t 1) t 3t 3 + t t + 1 t (ii) 4.9() 19.6 m (iii) st Q. 4. (i) Let t time Q is i motio. 1 d t + time P is i motio. s P s Q 47(t + ) 4.9(t + ) 64.6t 4.9t 47t t 19.6t t 4.9t t t s (ii) 64.6() 4.9() m Q. 5. First t secods u u, s 70, a 9.8, t t 70 ut 4.9t Equatio 1 First t secods u u, s , a 9.8, t t 10 ut 4.9 (t) 10 ut 19.6t Equatio Eq : 10 ut 19.6t Eq. 1: 140 ut + 9.8t 0 9.8t 00 98t t t 10 7 s

11 70 u ( 10 7 ) u 7 10 u 56 m/s Q. 6. (i) u u, a 9.8, s 30, t 5 s ut + at 30 5u + ( 9.8)(5) 9.5 5u u 18.5 m/s (ii) ( 9.8)(5) 30.5 Speed 30.5 m/s Q t 4.9t t t 16 t 10t (t )(t 8) 0 t, 8, t 8 + t 10 QED OR + t b a + t ( 10) 1 + t 10 Q t 4.9t t 49t,100 7t 100t Product of roots a c t t 300 QED Eercise E Q. 1. s d, t, u 0, a a s ut + at d a Equatio A Q.. (i) (ii) a d + k, t, u 0, a a s ut + at d + k a() d + k a Equatio B 4 Equatio A a 4d But a d + k d + k 4d 15ak k 3d QED 5k A B C 3k Distace i A (5k)(15ak) 37 ak Distace i C (3k)(15ak) ak Remaider 90ak 37 ak ak 30ak Time for B 30ak 15ak k s Total time 5k + k + 3k 10k s T t Let T the total time : t d : a 5a : 3a 5 8 : T ad t 3 8 T v 0 + (3a) ( 5 8 T ) 15 8 at Area 90ak (T) ( 15 8 at ) 90ak T 96k T 4 6 k 11

12 Q. 3. h ut + ( g)t h ut gt gt ut + h 0 Product of roots a c t h g QED Q. 4. Let PQ QR P Q R (i) The jourey P R: u u, v 7u, s, a a v u + as 49u u + (a)() 48u 4a 1u a a 1u The jourey P Q: u u, v v, s, a 1u v u + as v u + ( 1u )() v u + 4u v 5u v 5u (ii) P to Q: u u, v 5u, s, t, a a 5u u + a 4u a Q to R: u 5u, v 7u, t t, a a (ii) Distace A ( v a ) v v a Distace C ( v b ) v v Remaider s v b a v b Time take i B ( s v a v b ) v v s v a v b Total time take a v + ( v s v a v T v a + v b + v s t Let T the total time : t b : a b a + b : a a + b bt a + b, t at a + b v 0 + (a)( bt a + b ) abt a + b Area s ( (T) abt a + b ) s T s ( a + b ab ) QED Q. 6. (a) t 1, u u, a a, s s 1 s ut + at s 1 u( 1) + a( 1) b) + v b QED Q. 5. (i) 7u 5u + at t u a t QED u u + a a + a t, u u, a a, s s s ut + at s u + a 1 a A B C b Distace travelled, s s s 1 u + a a QED

13 Q. 7. (b) Whe, s u + a a u + 1 a 17 Whe 7, s u + 7a a u + 6 a 47 Solvig these gives a 6, u 8 (i) 10 s u + a a 8 + (6)(10) (6) 65 m (ii) s (6) (6 + 5) m (c) s + s ( + 1) Aswer: I the 0th ad 1st secods y z Average speed total distace total time v + yv + zv + y + z 5v 6 5v + 5yv + 5zv 3v + 6yv + 3zv 5 + 5y + 5z 3 + 6y + 3z + z y + z y Fractio of distace travelled at costat yv speed v + yv + zv y + y + z y + y + z y ( + z) + y y y + y y y 4 5 QED Q8 Q. 8. (i) 4 4 y z (ii) v 4 4z z ad v 4 Area uder curve d ( t y ) ( ((t y)) + y()(t y) + t y (t y) + yt y + (t y) d (t y) + yt y d t yt + y + yt y d )((t y)) d t y d y t y t d y t d QED + y t t y v 4 (t y) 13